How Can the Relative Speed of Photons Exceed the Cosmic Speed Limit?

[aleemudasir]

You've just repeated the same question. I already said this:


Just replace "missile" by "person" and the same thing applies. The whole point of relativity is this: all motion is relative. You want to know how I can tell if it is A that is moving towards me or me that is moving towards A. The answer is: it depends on the chosen frame of reference. There is no absolute frame of reference that will permit me to say that A is "really" moving and B is "really" at rest.

I repeated the question because in earlier case I had put a person in missile, and then you used that to answer me that the missile(in which the person was) is in rest w.r.t to the person, so now I just used two bodies(person A and B) to clarify what actually the question was, so as to tell you that nothing here is in rest w.r.t. anything so, how would reference frame be defined here.
Thanks.

 

[aleemudasir]

Then that equation doesn't apply because you're no longer working in an inertial frame (an inertial frame can't be moving at c).

OK what would be the case when one body is moving with 0.99999c(Body A) and other with c(Body B), and I am taking body A as reference frame?

 

[aleemudasir]

We can do it any way we want as long as the speed at which A and You are approaching calculates out to be the same in whatever frame we choose. So let's say that A and You are approaching at 0.8c.

We could pick a frame in which A is at rest and You are approaching from the +X direction at -0.8C.

Or we could pick a frame in which You are at rest and A is approaching from the -X direction at +0.8c.

Or we could pick a frame in which A and You are both moving at the same speed toward each other, A from -X at +0.5c and You from +X at -0.5c.

Or we could pick a frame in which A is moving from -X at +0.8c and You are moving from +X at +0.97561c.

So the frame we pick determines how we assign the motion(s) of the two persons.

Please note: I am only repeating what others have said before. We can't figure out why you are struggling so hard with this.

I just want to know few aspects of relative velocity taking relativity into account, and I think either I am not able to express my question or you aren't getting what I really want to know!

 

[Doc Al]

OK what would be the case when one body is moving with 0.99999c(Body A) and other with c(Body B), and I am taking body A as reference frame?

You'll find that the speed of Body B will be c as measured in any frame.

 

[elfmotat]

OK what would be the case when one body is moving with 0.99999c(Body A) and other with c(Body B), and I am taking body A as reference frame?

$$u=\frac{v+c}{1+vc/c^2}=\frac{v+c}{1+v/c}=c(\frac{v+c}{v+c})=c$$

So it doesn't matter what you use as v (the speed of body A), it will always come out as c. This should make sense because we expect light to have the same speed in every inertial frame.

 

[Michael C]

I repeated the question because in earlier case I had put a person in missile, and then you used that to answer me that the missile(in which the person was) is in rest w.r.t to the person, so now I just used two bodies(person A and B) to clarify what actually the question was, so as to tell you that nothing here is in rest w.r.t. anything so, how would reference frame be defined here.
Thanks.

You can still define a reference frame to be at rest with respect to A, or one that is at rest with respect to B. You can also define a reference frame to be moving at a velocity x with respect to A, or one that moves at velocity y with respect to A, or one that moves at velocity z with respect to B...

In short, you can define an infinity of reference frames for any situation. Each one is as valid as the next one.

 

[ghwellsjr]

We can do it any way we want as long as the speed at which A and You are approaching calculates out to be the same in whatever frame we choose. So let's say that A and You are approaching at 0.8c.

We could pick a frame in which A is at rest and You are approaching from the +X direction at -0.8C.

Or we could pick a frame in which You are at rest and A is approaching from the -X direction at +0.8c.

Or we could pick a frame in which A and You are both moving at the same speed toward each other, A from -X at +0.5c and You from +X at -0.5c.

Or we could pick a frame in which A is moving from -X at +0.8c and You are moving from +X at +0.97561c.

So the frame we pick determines how we assign the motion(s) of the two persons.

Please note: I am only repeating what others have said before. We can't figure out why you are struggling so hard with this.

I just want to know few aspects of relative velocity taking relativity into account, and I think either I am not able to express my question or you aren't getting what I really want to know!

This thread started off with a question about two photons approaching each other. Unfortunately, that question has several different unrelated issues with regard to it and they have all been covered in different posts. If the question had been regarding two bodies (or persons or missiles or whatever as long as they are traveling less than the speed of light), then the answers could have been focused on that one issue. Unfortunately, you continue to bring in one body traveling at the speed of light and it diverts the answers off in another direction again. So please don't bring up a body traveling at the speed of light, OK, it will just bring about more confusion. A body can go at any speed approaching the speed of light such as 0.999999999c but not 1.0000000000c.

Now in my previous answer, I picked a relative speed of 0.8c to use in my examples. Note that I said we could use a reference frame in which both persons are approaching at 0.5c. If you simply added these two speeds together, you would get exactly 1.0c. Do you understand that in this reference frame, even though the closing speed is 1.0c, there is nothing traveling at 1.0c and the two bodies don't have a relative speed of 1.0c? If you use the velocity addition formula, you will see that their relative speed is 0.8c. Do you understand all this?

Whether or not this is what you want to know, do you understand what I have explained? If it isn't what you want to know, tell me what about my explanation is not related to what you want to know.

 

[harrylin]

This comes back every now and then; isn't it a FAQ?

I understand that the speed of any of the two massive objects w.r.t observer will be 0.999c but when we see from frame of reference of any of the moving objects the speed of the other object w.r.t it will be 1.98c which is more than the cosmic limit.
P.S. As far as I have understood speed is always relative(correct me if I am wrong).

The limit for that kind of relative speed is simply the sum of each: thus c+c=2c. And why this is so, has been elaborately explained already; that can't be the problem. The only problem is to correct the misunderstanding - and that can be a big problem, if the misunderstanding has been ingrained.

Apparently you think that a light ray can only move with a velocity of c relative to another object or light ray. So, here's another reply -by a certain A. Einstein- in one of the papers in which also your "cosmic limit" is explained:

"the ray moves relatively to the [moving] initial point of k, when measured in the stationary system, with the velocity c-v".
- http://www.fourmilab.ch/etexts/einstein/specrel/www/

Note that c-v>c for the case that the two move in opposite directions.

Just now want to know one thing.Is the relative speed of one photon wrt other is greater than c or not?If not then why and how the distance is shrinking the other way?

That question is also answered in that same section; as long as you use a single reference system you must subtract the two velocities from each other, taking in account the angles. Thus for perpendicular motion one has (a few lines above the earlier quote):

"applied to the [moving] axes of Y and Z—it being borne in mind that light is always propagated along these axes, when viewed from the stationary system, with the velocity √(c²-v²)"

That's standard Pythagoras.

 

[aleemudasir]

$$u=\frac{v+c}{1+vc/c^2}=\frac{v+c}{1+v/c}=c(\frac{v+c}{v+c})=c$$

So it doesn't matter what you use as v (the speed of body A), it will always come out as c. This should make sense because we expect light to have the same speed in every inertial frame.

Thanks, but!
If both the bodies are moving in the same direction in the +ve x-direction the value of relative velocity comes up:
$$u=\frac{v-c}{1-vc/c^2}=\frac{v-c}{1-v/c}=c(\frac{v-c}{c-v})=-c$$

What does -c mean here?

 

[aleemudasir]

This thread started off with a question about two photons approaching each other. Unfortunately, that question has several different unrelated issues with regard to it and they have all been covered in different posts. If the question had been regarding two bodies (or persons or missiles or whatever as long as they are traveling less than the speed of light), then the answers could have been focused on that one issue. Unfortunately, you continue to bring in one body traveling at the speed of light and it diverts the answers off in another direction again. So please don't bring up a body traveling at the speed of light, OK, it will just bring about more confusion. A body can go at any speed approaching the speed of light such as 0.999999999c but not 1.0000000000c.

Now in my previous answer, I picked a relative speed of 0.8c to use in my examples. Note that I said we could use a reference frame in which both persons are approaching at 0.5c. If you simply added these two speeds together, you would get exactly 1.0c. Do you understand that in this reference frame, even though the closing speed is 1.0c, there is nothing traveling at 1.0c and the two bodies don't have a relative speed of 1.0c? If you use the velocity addition formula, you will see that their relative speed is 0.8c. Do you understand all this?

Whether or not this is what you want to know, do you understand what I have explained? If it isn't what you want to know, tell me what about my explanation is not related to what you want to know.

I understood the difference between closing speed and relative speed, and how the relative speed comes up 0.8c!

 

[Doc Al]

Thanks, but!
If both the bodies are moving in the same direction in the +ve x-direction the value of relative velocity comes up:
$$u=\frac{v-c}{1-vc/c^2}=\frac{v-c}{1-v/c}=c(\frac{v-c}{c-v})=-c$$

What does -c mean here?

It just means you messed up with a sign.

You'll have less chance of messing up signs when you use this version of the addition of velocity formula:
$$V_{a/c} = \frac{V_{a/b} + V_{b/c}}{1 + (V_{a/b} V_{b/c})/c^2}$$

Here $V_{a/c}$ means the velocity of A with respect to C. So call the first body C and the second body A. And the third frame B.

So V_a/b = c;
V_c/b = v, so V_b/c = -v;

What you want is u = V_a/c.

Do all that and you'll get u = +c.

 

[ghwellsjr]

Thanks, but!
If both the bodies are moving in the same direction in the +ve x-direction the value of relative velocity comes up:
$$u=\frac{v-c}{1-vc/c^2}=\frac{v-c}{1-v/c}=c(\frac{v-c}{c-v})=-c$$

What does -c mean here?

You need to substitue v with -c instead of putting the minus sign in front of the c. Then you will get:

$$u=\frac{-c+c}{1-cc/c^2}=\frac{-c+c}{1-1}=\frac{0}{0}=?$$

which is indeterminate meaning you need some other way to calculate the answer.

However if you start with the orginal equation:

$$u=\frac{v+u'}{1+vu'/c^2}$$

and set u' equal to -v (since the formula assumes opposite directions and you want the same direction), then:

$$u=\frac{v-v}{1-vv/c^2}=\frac{0}{1-v^2/c^2}=0$$

unless v=c but u approaches zero in the limit as v approaches c.

 

[ghwellsjr]

This thread started off with a question about two photons approaching each other. Unfortunately, that question has several different unrelated issues with regard to it and they have all been covered in different posts. If the question had been regarding two bodies (or persons or missiles or whatever as long as they are traveling less than the speed of light), then the answers could have been focused on that one issue. Unfortunately, you continue to bring in one body traveling at the speed of light and it diverts the answers off in another direction again. So please don't bring up a body traveling at the speed of light, OK, it will just bring about more confusion. A body can go at any speed approaching the speed of light such as 0.999999999c but not 1.0000000000c.

Now in my previous answer, I picked a relative speed of 0.8c to use in my examples. Note that I said we could use a reference frame in which both persons are approaching at 0.5c. If you simply added these two speeds together, you would get exactly 1.0c. Do you understand that in this reference frame, even though the closing speed is 1.0c, there is nothing traveling at 1.0c and the two bodies don't have a relative speed of 1.0c? If you use the velocity addition formula, you will see that their relative speed is 0.8c. Do you understand all this?

Whether or not this is what you want to know, do you understand what I have explained? If it isn't what you want to know, tell me what about my explanation is not related to what you want to know.

I understood the difference between closing speed and relative speed, and how the relative speed comes up 0.8c!

So are we done?

 

[aleemudasir]

It just means you messed up with a sign.

You'll have less chance of messing up signs when you use this version of the addition of velocity formula:
$$V_{a/c} = \frac{V_{a/b} + V_{b/c}}{1 + (V_{a/b} V_{b/c})/c^2}$$

Here $V_{a/c}$ means the velocity of A with respect to C. So call the first body C and the second body A. And the third frame B.

So V_a/b = c;
V_c/b = v, so V_b/c = -v;

What you want is u = V_a/c.

Do all that and you'll get u = +c.

Sorry for that question, I realized just a few moments after posting the question that I have done a mistake while solving the equation!
Anyways thanks!

 

[aleemudasir]

So are we done?

Yes to some extent, now I understood the dilemma of this relative velocity taking relativity into account. I have some last questions regarding relativity, do we know the reason why a particle moving @c can't be taken as reference frame and why does the behavior/properties and laws change so abruptly when we move from speed c-1m/s to c?

Thanks to all for you kind help and precious time.
With regards
Mudasir

 

[harrylin]

Yes to some extent, now I understood the dilemma of this relative velocity taking relativity into account. I have some last questions regarding relativity, do we know the reason why a particle moving @c can't be taken as reference frame and why does the behavior/properties and laws change so abruptly when we move from speed c-1m/s to c? [..]

There is no "abrupt" change. Instead, at speed c such a reference frame (even a virtual one) would have (according to us) clocks that do not tick at all (dt'=0) and its rulers would have zero length in the direction of motion; you can't calculate with that.

 

[ghwellsjr]

Yes to some extent, now I understood the dilemma of this relative velocity taking relativity into account. I have some last questions regarding relativity, do we know the reason why a particle moving @c can't be taken as reference frame and why does the behavior/properties and laws change so abruptly when we move from speed c-1m/s to c?

Thanks to all for you kind help and precious time.
With regards
Mudasir

No matter how much you have accelerated in the past to achieve some high rate of speed, even c-1m/s, the speed of light to you will still measure to be c. But remember, we're talking about measuring the time it takes for light to make a round trip and then dividing that by 2 and assuming that the one-way speed of light is equal to that answer.

 

[aleemudasir]

No matter how much you have accelerated in the past to achieve some high rate of speed, even c-1m/s, the speed of light to you will still measure to be c. But remember, we're talking about measuring the time it takes for light to make a round trip and then dividing that by 2 and assuming that the one-way speed of light is equal to that answer.

I think you got my point wrong, when I said moving from c-1m/s to c I didn't mean accelerating from c-1m/s to c rather I meant when we observe/study one object which is moving at c-1m/s and another object moving at c.
Thanks

 

[ghwellsjr]

I think you got my point wrong, when I said moving from c-1m/s to c I didn't mean accelerating from c-1m/s to c rather I meant when we observe/study one object which is moving at c-1m/s and another object moving at c.
Thanks

If there's one thing you should have gotten by now in this thread, it's that no object can move at c.

 

[Dale]

do we know the reason why a particle moving @c can't be taken as reference frame

Yes, it leads to a self contradiction.
https://www.physicsforums.com/showthread.php?t=511170

 

[Doc Al]

and another object moving at c.

This second 'object' must be massless... such as a photon. Not any kind of ordinary 'object'.