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Seeing things in their past? You are full of beans!

Polymath257

Think & Care
Staff member
Premium Member
Would an example of two ends of a light path, be from star to earth? Please explain how you have two events(involving light) on that light path, that are both in the same reference frame.

The two events are the place and time when the light starts its journey (the first event) and the place and time when it finishes its journey (the second event). Events always have both location and time. So, the two events here would be the start of the journey (location and time at the earth) and the end of the journey (location and time at the star).

Your question about the reference frame makes no sense. For example, the earth determines where and when the first is (using clocks, etc) and then where and when the second is. The distance between them *as measured by the Earth* is x and the time between them *as measured by the Earth* is t.

If the star is at rest with respect to the earth, then someone at the star would measure the same distance, x, and the same time of travel, t. The star and the earth are in the *same frame* because they are at rest with respect to each other.

Once again, it seems that you don't quite get what a reference frame is. ALL that is required to be a reference frame is that you not be experiencing acceleration. So the sun is very nearly in a reference frame (the only acceleration is due to its motion around the center of the galaxy, which is a very small acceleration). The Earth isn't as good because of its much faster motion around the sun, but can be used in many situations (and we will, for convenience).

So, two different reference frames are not accelerating with respect to each other: in other words, they are moving with some constant speed with respect to each other. The speed is called the relative speed.

Each reference frame can use its clocks, measuring sticks, trigonometry, etc to determine the locations and times of *all* events. So, the earth can measure when a beam of light leaves the earth and also when it gets to the sun (or star, if you prefer). Those are two events. The distance between them *as measured by the Earth* and the time between them *as measured by the earth* are the distance and time between them in the reference frame of the earth. So, *in the frame of the earth*, the distance between the earth and the sun is 93 million miles and the time it takes for light to go from the earth to the sun is 500 seconds. Those are measurements *in the reference frame of the earth*.

Similarly, the earth can measure the distance to a star and the time it takes for some spacecraft (or light) to go from the earth to that star. Those are measurements *in the reference frame of the earth*.

Now, one basic aspect of reference frames is that you are always at rest in your own reference frame (by definition of what it means to be your reference frame). So, if I am traveling past the earth, and someone on earth measures me to be going at 50,000 km/s, that is my speed in the reference frame of the earth. In my own reference frame, I am *at rest*. The earth is going past *me* at 50,000 km/s.

So, suppose we send a beam of light from the earth to the sun. In the reference frame of the earth, the distance the light travels is 93 million miles and the time it takes is 500 seconds. Two events: one on the earth when the light starts out, the other on the sun when the light gets there, separated by both distance and time.

Now, from the reference frame of someone going past the earth at .5c, the distance and the time of that journey of the light will be different.There are still two events: the place and time when the light left the earth and the place and time when it gets to the sun. But the *distance* between those two events will NOT be 93 million miles. And the time it takes for light to go between those two events will NOT be 500 seconds. If that someone is moving towards th sun from the earth, then the distance between those two events will be about 53 million miles and the time will be 288 seconds.

The speed of light is still the same in this other reference frame (186, 000 miles per second). The other reference frame can still measure distances and times for *all* events. But the measured distances and times will be different than those measured by someone on the earth (because of the relative speeds).

The events themselves are NOT in a specific frame: it is the distance and time between those events that are in some frame. ALL frames can measure distance and times between those two events. ALL frames can compute the speed of light by taking the distance *as measured* by the time *as measured* and all will get the same answer: 186,000 miles per second.

Similarly, *every* reference frame will compute the same proper time between two events. They *all* compute that proper time by doing
sqrt( t^2 - x^2 /c^2 )

Each frame has its own values for x and t, but ALL will obtain the same value for the proper time. PLUS, that proper time is what the frame measures for the time that sees itself as *at rest* while going between the events. In a frame that measures x and t above, the 'rest frame' is moving at v=x/t.
 

Polymath257

Think & Care
Staff member
Premium Member
Isn't that what I did in my example? Observer A, measures the time it takes for light to get from star to earth. Observer B is in a ship moving at speed v with respect to observer A and is the one measuring the proper time. So when his ship reaches v=c, he measures proper time=0

No. Observer A is measuring the time it takes for *B* to go from the earth to the sun at speed v. The proper time is the time *as measured by B*. So, we are watching the time that *B* experiences as the speed of *B* with respect to the earth gets closer and closer to c. This is the limit I am taking here (although it isn't required in the other formulation).

What are you calling two events in this example, are you meaning the two different observations?

The two events are when the ship leaves the earth (place and time for first event) and when the ship reaches the sun (place and time for the second event).
 

Bird123

Well-Known Member
We observe an object travelling towards the Earth that is 1ly away . The object is travelling at 0.5c towards us.

The object takes 2ly to arrive, if we were seeing that object in the past, it would not arrive in 2ly, proving ''you'' are full of beans and are incompetent at ''your'' job.


What you fail to realize is that time is not constant. Your time and the objects time are not the same. Your eyes are light receptors and your brain translates this light into an image for you. Since, the light took 2 years to arrive at your eyes, technically , you are looking into the past. As the object gets closer, the distance is shorter and there is no long the time lag. Your time view will seamlessly catch up. Time rates mesh seamlessly together since your eyes can not see time.

When will the object really reach you and how long was it traveling before you saw it? There are many many variables involved. How many gravity wells altered the object's path, speed, and time?

I know one must always look beyond the surface. Knowledge and complexity almost always lives beyond that simple surface.

Hopes this helps some.
 

TrueBeliever37

Well-Known Member
No. Observer A is measuring the time it takes for *B* to go from the earth to the sun at speed v. The proper time is the time *as measured by B*. So, we are watching the time that *B* experiences as the speed of *B* with respect to the earth gets closer and closer to c. This is the limit I am taking here (although it isn't required in the other formulation).



The two events are when the ship leaves the earth (place and time for first event) and when the ship reaches the sun (place and time for the second event).

These last two posts you made at least helped me understand better what you are saying.
 

TrueBeliever37

Well-Known Member
What you fail to realize is that time is not constant. Your time and the objects time are not the same. Your eyes are light receptors and your brain translates this light into an image for you. Since, the light took 2 years to arrive at your eyes, technically , you are looking into the past. As the object gets closer, the distance is shorter and there is no long the time lag. Your time view will seamlessly catch up. Time rates mesh seamlessly together since your eyes can not see time.

When will the object really reach you and how long was it traveling before you saw it? There are many many variables involved. How many gravity wells altered the object's path, speed, and time?

I know one must always look beyond the surface. Knowledge and complexity almost always lives beyond that simple surface.

Hopes this helps some.

Hi Bird123,

I think the actual time itself is constant, it's just the measurement of that time, that is different.

If a ship was traveling to us, and we didn't see it until 2 years after it left, we aren't seeing a ship 2 years into the past. We are just seeing a ship that took 2 years to get here.
 

Polymath257

Think & Care
Staff member
Premium Member
Hi Bird123,

I think the actual time itself is constant, it's just the measurement of that time, that is different.

If a ship was traveling to us, and we didn't see it until 2 years after it left, we aren't seeing a ship 2 years into the past. We are just seeing a ship that took 2 years to get here.

Think of it like this. Suppose that ship was carrying a message. How old is the message when the ship gets to you?

In the case of light from another star, the light carries information about the star (color, for example) and takes years to get to you. So that information was about that star some years ago. What we see now is what the star was like when the light started out.
 

TrueBeliever37

Well-Known Member
Think of it like this. Suppose that ship was carrying a message. How old is the message when the ship gets to you?

In the case of light from another star, the light carries information about the star (color, for example) and takes years to get to you. So that information was about that star some years ago. What we see now is what the star was like when the light started out.

Here's what I just can't get.

If t=0, and d=0 for photons, then WHY does it take longer in every valid frame of reference, for light to get here from 30 million light years away, than from the Sun to the earth? I mean the measured distances and times may change in other reference frames, but there is always a difference.

There has to be some reason why in every valid frame, there is a difference in time involved to cover the two different distances.

If time and distance were truly 0 for the photon, then it looks like the photon could travel from it's source to any other point in space instantly. Which would then mean we shouldn't see a time difference when using our frame of reference.
 
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TrueBeliever37

Well-Known Member
Think of it like this. Suppose that ship was carrying a message. How old is the message when the ship gets to you?

In the case of light from another star, the light carries information about the star (color, for example) and takes years to get to you. So that information was about that star some years ago. What we see now is what the star was like when the light started out.

Right, but why is it taking years to get to us, if t=0 and d=0 for the light?
 

Polymath257

Think & Care
Staff member
Premium Member
Here's what I just can't get.

If t=0, and d=0 for photons, then WHY does it take longer in every valid frame of reference, for light to get here from 30 million light years away, than from the Sun to the earth? I mean the measured distances and times may change in other reference frames, but there is always a difference.

Well, the light itself doesn't have a reference frame. We can compute the proper time, but that has little relation to coordinate time (which is always in a reference frame).

There has to be some reason why in every valid frame, there is a difference in time involved to cover the two different distances.

Because there is no rest frame moving at the speed of light.

If time and distance were truly 0 for the photon, then it looks like the photon could travel from it's source to any other point in space instantly. Which would then mean we shouldn't see a time difference when using our frame of reference.

And this is wrong. The proper time being 0 doesn't mean any coordinate time has to be 0. They are not directly related concepts.
 

Bird123

Well-Known Member
Hi Bird123,

I think the actual time itself is constant, it's just the measurement of that time, that is different.

If a ship was traveling to us, and we didn't see it until 2 years after it left, we aren't seeing a ship 2 years into the past. We are just seeing a ship that took 2 years to get here.


Time is not constant. Did you know the time settings on all those TV satellites must be adjusted all the time? If left alone, your TV show would not show up when it was supposed to. They use the same measurement tools on both.

You are right in that seeing the ship is not a current past. It is the image of what happened it the past. Let's work on another example. If you had a switch which could turn the sun off, do you know what would happen? When you turned the sun off, you would never know it was done until about 8 minutes after it was done. It takes that long for it all to get here. Time is not constant. Light speed is limited. Your eyes see nothing until the light hits them.

I know it might all sound crazy but things would not work any other way.
 

TrueBeliever37

Well-Known Member
Time is not constant. Did you know the time settings on all those TV satellites must be adjusted all the time? If left alone, your TV show would not show up when it was supposed to. They use the same measurement tools on both.

You are right in that seeing the ship is not a current past. It is the image of what happened it the past. Let's work on another example. If you had a switch which could turn the sun off, do you know what would happen? When you turned the sun off, you would never know it was done until about 8 minutes after it was done. It takes that long for it all to get here. Time is not constant. Light speed is limited. Your eyes see nothing until the light hits them.

I know it might all sound crazy but things would not work any other way.

The fact that it takes a little more than 8 minutes for the light to get here, is what lets me know time can't equal 0 for light. If the time and distance were truly 0 for the photons, then they would arrive instantly. It wouldn't take us 500s to notice it. And we wouldn't be able to tell a difference in time, we wouldn't know whether the light was from the Sun or from another star 30 million light years away.
 

Polymath257

Think & Care
Staff member
Premium Member
The fact that it takes a little more than 8 minutes for the light to get here, is what lets me know time can't equal 0 for light. If the time and distance were truly 0 for the photons, then they would arrive instantly. It wouldn't take us 500s to notice it. And we wouldn't be able to tell a difference in time, we wouldn't know whether the light was from the Sun or from another star 30 million light years away.

The fact that the proper time,
sqrt( t^2 - x^2 /c^2 )

is 0 does not mean that t=0. In fact, it *prevents* t=0 unless x=0 (which means we are only talking about one event).

You are confusing coordinate time (500 seconds, in the frame of the Earth) and proper time ( always 0 for light). They are not the same! Use the equation above to see the difference.
 

TrueBeliever37

Well-Known Member
Well, the light itself doesn't have a reference frame. We can compute the proper time, but that has little relation to coordinate time (which is always in a reference frame).



Because there is no rest frame moving at the speed of light.



And this is wrong. The proper time being 0 doesn't mean any coordinate time has to be 0. They are not directly related concepts.

That still doesn't explain why we are able to see a difference in time here, since the photon should have been able to travel between any 2 points in space in no time flat.

If time and distance are 0 for the photons, then WHY are we able to see a difference in time when light travels here from the Sun,versus when it travels here from another Star?
 

TrueBeliever37

Well-Known Member
The fact that the proper time,
sqrt( t^2 - x^2 /c^2 )

is 0 does not mean that t=0. In fact, it *prevents* t=0 unless x=0 (which means we are only talking about one event).

You are confusing coordinate time (500 seconds, in the frame of the Earth) and proper time ( always 0 for light). They are not the same! Use the equation above to see the difference.

Then that should mean that it really does take time for the photon to travel different expanses in space. (If proper time =0 does not mean t=0)
 

gnostic

The Lost One
That still doesn't explain why we are able to see a difference in time here, since the photon should have been able to travel between any 2 points in space in no time flat.
Who said that? That it travel through space “in no time flat”?

You? Or scientists?
 

TrueBeliever37

Well-Known Member
Who said that? That it travel through space “in no time flat”?

You? Or scientists?
If t=0 and d=0 for a photon to travel, isn't that what is implied? No matter if it is traveling what we would call 30 million light years or 500 light seconds, the photon is said to experience no time or distance.
 

Polymath257

Think & Care
Staff member
Premium Member
That still doesn't explain why we are able to see a difference in time here, since the photon should have been able to travel between any 2 points in space in no time flat.

If time and distance are 0 for the photons, then WHY are we able to see a difference in time when light travels here from the Sun,versus when it travels here from another Star?

Once again, there is no reference frame for the photon. Talking about coordinate distances and time only makes sense in a reference frame. The closest you can get for a photon is for proper time, which is always zero for a photon. There is no conversion factor going from proper time to coordinate time.

You are trying to do a conversion (proper time to coordinate time) that makes no sense.

We see differences in time: those are coordinate time. Those are NOT directly related to proper time for the photon path, which is all that makes sense in the case of the photon. NO CONVERSION FACTOR RELATES PROPER TIME AND COORDINATE TIME!
 

Polymath257

Think & Care
Staff member
Premium Member
If t=0 and d=0 for a photon to travel, isn't that what is implied? No matter if it is traveling what we would call 30 million light years or 500 light seconds, the photon is said to experience no time or distance.

Were trying to make *some* sense of your question of what the photon 'experiences'. Since there is no coordinate time for the photon ( it has no rest frame), the only sensible answer is the proper time.

So, no, that is NOT what it means. For a photon, any motion has to be in *a different* frame, since it has no frame.

One alternative is to simply say your question about how much time the photon experiences makes no sense: no reference frame exists, so no answer is possible.

The *only* way to make sense of the question is to use proper time, which has NO implications for 'motion' in some frame.
 

TrueBeliever37

Well-Known Member
Were trying to make *some* sense of your question of what the photon 'experiences'. Since there is no coordinate time for the photon ( it has no rest frame), the only sensible answer is the proper time.

So, no, that is NOT what it means. For a photon, any motion has to be in *a different* frame, since it has no frame.

One alternative is to simply say your question about how much time the photon experiences makes no sense: no reference frame exists, so no answer is possible.

The *only* way to make sense of the question is to use proper time, which has NO implications for 'motion' in some frame.


Then what does it even mean to say t=0, and d=0 for a photon? It makes no sense to say it travels at c, yet time and distance are both 0.
 
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