Implications of Einstein's Theories

[jtbell]

So to go back to how I kept mistakenly visualizing this, if I was following behind a photon at .9c then it would actually appear to me to be moving away from me at .1c, correct? This seems to be unrelated to relativity because this example does not involve reference frames or any observers.

To add onto Drakkith's answer, have you seen the "relativistic velocity addition" equation yet?

http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/einvel.html
http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/einvel2.html

Try the different examples, with the "projectile" being a photon, then see if you can set up the situation you describe above. You should find that observers A and B always both find the photon's velocity to be c.

 

[JimiJams]

I was trying to create a scenario where there is no other observer than myself. Imagine I'm on an empty train and I'm trailing behind a photon at .9c. You're saying that photon, in this situation where I am the only observer, will still appear to move away at c? If that's the case then there must be two relative reference frames for the math to work, how would we define them?

 

[jtbell]

I'm trailing behind a photon at .9c.

This implies an external observer, or more precisely, another reference frame besides the one in which you are at rest. You are moving at .9c with respect to whom or what?

With respect to yourself, you are stationary.

 

[PAllen]

I was trying to create a scenario where there is no other observer than myself. Imagine I'm on an empty train and I'm trailing behind a photon at .9c. You're saying that photon, in this situation where I am the only observer, will still appear to move away at c? If that's the case then there must be two relative reference frames for the math to work, how would we define them?

Traveling at .9c relative to what? It can't be relative to a photon, because a photon moves at c relative to all trains.

 

[JimiJams]

Let's say me and the photon are both moving above a stationary surface. I'm not in any flying object, my body is just moving. I'm traveling .9c relative to the stationary ground, and the photon is moving c relative to the ground. There are checkmarks on the ground indicating a certain distance, and I had two stopwatches. Every time the photon passes a checkmark I hit the button to record at what point past t=0 the photon passes that checkmark. I also hit my other stopwatch every time I pass a check mark.

At the end wouldn't I be able to calculate my speed at .9c and the photon's at c, showing that the photon is really only moving .1c relative to me?

If this is not the case, then where would I assign the two reference frames?

 

[WannabeNewton]

No because you're using the Galilean conception of relative velocity. In Galilean relativity, if a particle $O$ has velocity $\vec{v}$ relative to an inertial frame and another particle $O'$ has velocity $\vec{v}'$ relative to the same inertial frame then $O'$ has a velocity $\vec{v}' - \vec{v}$ relative to $O$. This is the formula you're trying to apply in your scenario but you cannot do this because Galilean addition of velocities is only accurate to 1st order in $\frac{v^2}{c^2}$ when $v << c$.

When you have $v\sim c$ you need to use the special relativistic velocity addition formula: http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/einvel.html

 

[Drakkith]

At the end wouldn't I be able to calculate my speed at .9c and the photon's at c, showing that the photon is really only moving .1c relative to me?

No, as you aren't measuring the photons velocity relative to yourself, but to the ground. You would be measuring the rate that the distance between yourself and the photon increases as seen from the stationary ground.

Imagine you had a very long set of distance markers stretched out in front of you, with 1 meter between each marker. An observer on the ground also has markers set up, with 1 meters in between each of them. You are moving at 0.866c relative to the observer on the ground. In one second, as measured by you, the photon will pass 300 million of your markers. To the observer on the ground, the photon passes 300 million of his markers per second according to his watch.

 

[ghwellsjr]

I was trying to create a scenario where there is no other observer than myself. Imagine I'm on an empty train and I'm trailing behind a photon at .9c. You're saying that photon, in this situation where I am the only observer, will still appear to move away at c? If that's the case then there must be two relative reference frames for the math to work, how would we define them?

Here's what I think you want. I'm going to show you stationary in your own rest frame and you emit a photon. I'm defining light to travel at 1000 feet per microsecond (usec). You are shown as the thick black line with dots every microsecond. The photon is defined (not measured, not observed, not seen) to travel at 1000 feet per microsecond in any Inertial Reference Frame (IRF) and is shown as the thin black line. After 5 microseconds we (not you) see that the photon is 5000 feet in front of you:

Next we transform (using the Lorentz Transformation process as I described in post #25) this scenario to one in which you are traveling at 0.9c (by using β=-0.9c), chasing the photon:

Now we (not you) see that the photon is just 1000 feet in front of you after 10 microseconds.

Do you understand why I repeatedly said that we and not you can see the propagation of the photon?

[NOTE: I made a mistake in post #25 where I said that "it takes 3 nanoseconds for each flash to make the round trip". I should have said "it takes 6 nanoseconds".]

 

[ghwellsjr]

Let's say me and the photon are both moving above a stationary surface. I'm not in any flying object, my body is just moving. I'm traveling .9c relative to the stationary ground, and the photon is moving c relative to the ground. There are checkmarks on the ground indicating a certain distance, and I had two stopwatches. Every time the photon passes a checkmark I hit the button to record at what point past t=0 the photon passes that checkmark. I also hit my other stopwatch every time I pass a checkmark.

At the end wouldn't I be able to calculate my speed at .9c and the photon's at c, showing that the photon is really only moving .1c relative to me?

If this is not the case, then where would I assign the two reference frames?

First, you should realize, as I have repeatedly said, you cannot see the photons. You cannot tell when they hit the checkmarks. All you can do is wait for the light to reflect back from the checkmarks and then you can measure the roundtrip time it took for the light to leave you (back at time zero) propagate to a checkmark and reflect back to you. You have no idea where during that roundtrip it hit the checkmark.

But what you do if you want to follow the precepts of Special Relativity, is you assume that the light took exactly the same amount of time to propagate to the checkmark as it took for the reflected light to propagate back to you. And you also assume that the light propagated at c in both directions. This allows you to construct your own reference frame in which you are at rest.

Let's see how this works starting with the second diagram that I drew from my previous post except this time I've added in two checkmarks every 5000 feet. I've shown the light that reflects back to you and you can determine when your stopwatches would measure both the roundtrip light signals and when you pass the checkmarks:

It looks to me like you see the light from the first checkmark at your time of 2.3 usecs and the second one at 4.6 usecs. You divide these times in half to assign the times that light reflected off the checkmarks and you calculate how far light travels in that time to determine how far away they were from you at that time. So you would determine that the first checkmark was 1150 feet away from you at your time of 1.15 usecs and the second checkmark was at 2300 feet away at your time of 2.3 usecs.

It also looks to me like you passed the first checkmark at your time of 2.4 usecs and the second one at 4.8 usecs.

Now you have all the information you need to calculate the speeds of the checkmarks relative to you.

For the first checkmark, you determined that at your time of 1.15 usecs, it was 1150 feet away and then at 2.4 usecs, it was 0 feet away. Therefore, it traveled 1150 feet in (2.4-1.15) = 1.25 usecs for a speed of 1150/1.25 or 920 feet per usec. The actual speed is 900 feet per usec, so this is pretty good for eyeballing.

For the second checkmark, you determined that at your time of 2.3 usecs, it was 2300 feet away and then at 4.8 usecs, it was 0 feet away. Therefore, it traveled 2300 feet in (4.8-2.3) = 2.5 usecs for a speed of 2300/2.5 or 920 feet per usec, same as for the first checkmark.

Finally, for the speed of the photons, as I said before, you simply assume that they travel at c. That's what you did to determine the speed of the checkmarks. There is no way that you can actually measure their speed apart from assuming their speed to begin with.

So here is the frame that you construct from the measurements you took and the assumption you made about the speed of light and the calculations that you made:

This is exactly the same diagram that you get by transforming the above frame to a speed of 0.9c (in fact, that's just how I got it).

 

[Simon Bridge]

Let's say me and the photon are both moving above a stationary surface. I'm not in any flying object, my body is just moving. I'm traveling .9c relative to the stationary ground, and the photon is moving c relative to the ground. There are checkmarks on the ground indicating a certain distance, and I had two stopwatches. Every time the photon passes a checkmark I hit the button to record at what point past t=0 the photon passes that checkmark. I also hit my other stopwatch every time I pass a checkmark.

You have to say how you know where the photon is.

Lets modify this experiment a bit.
Instead of following a single photon, you are following a well-located pulse of light.
The pulse will spread out as it travels so we'll keep the experiment short enough in duration that this is not a problem.

We will define two reference frames -
A: the rest-frame of the ground. Observer - "Alice"
B: the rest-frame of you, "Bob", which has relative velocity of v wrt tA "Alice" and the ground.

The light travels at speed c in both frames.

At regular (say: 1 light-second, as measured by Alice) intervals there are light detectors stationary wrt Alice. When the pulse maximum intensity is detected, they emit a flash of light. In this way Bob and Alice can track the progress of the same pulse as it speeds away from them.

This should avoid a bunch of objections.
There are still a few technical considerations, but they are mainly engineering problems.

At the end wouldn't I be able to calculate my speed at .9c and the photon's at c, showing that the photon is really only moving .1c relative to me?

You will always calculate your own speed as zero.
This is where you have to be careful in your language - when you want to work out a speed, you have to say who is doing the observing: what is the speed relative to? If you don't say, then the convention is that it is relative to yourself (or the last observer). Your speed relative to yourself is always zero.

You could work out that Alice would calculate that you were doing 0.9c and that the light pulse is moving at 0.1c relative to you. That is correct.
But so what?

There is a habit you get into with Newtonian mechanics, where you pretend that something is "really" stationary and everything else moves. We end up thinking that there really is such a thing as absolute rest. In Galilean relativity that idea looks a bit shaky but Special relativity kills it stone dead.

If you want to say that you are "really" traveling at 0.9c, then you must be saying that Alice's POV is somehow special. But why pick hers? Why not someone elses? The choice is arbitrary.


Take a more extreme example:
If you and I were to head at 0.9c to A, in opposite directions, then Alice would calculate our relative velocity to each other as 1.8c. Either of us could calculate that Alice would calculate that but, again, so what?
What doe that mean to either of us?