# Difficulty Understanding Special Relativity Time Dilation [closed]

Joe and I have agreed to shoot a photon at each other when our clocks count 10 seconds. Suppose I am approaching Joe at relativistic speeds such that I think Joe’s clock is ticking at half my clock’s speed.

So, I fire my photon when his clock says 5 (and the same for him, reversed). Yet, if I’m only 4 light seconds away at this time, I can’t figure out why I won’t see the photon reach him before I see his clock reaches 10 seconds despite that being impossible.

Is the problem just that we couldn’t start our clocks at the same time?

• Please clarify what you mean. What is the initial distance? In which frame are the clocks synchronized? Commented Jul 31 at 21:20
• "when our clocks count ten seconds" --- ten seconds from what? Commented Jul 31 at 21:27
• en.wikipedia.org/wiki/Spacetime_diagram I would strongly recommend you draw a spacetime diagram of the situation you're investigating, and if you draw one in your frame and one in Joe's frame, the answer to your question should be clear. Commented Jul 31 at 21:34
• You need to specify everything clearly in relativity so people know exactly what you're asking, and so you don't confuse yourself. ;) In particular, you need to be aware of the distinction between what you see vs what you measure after taking delays due to the finite speed of light into account. You may find it helpful to look at some of robphy's spacetime diagrams on rotated graph paper, eg physics.stackexchange.com/a/383363/123208 & physics.stackexchange.com/a/319203/123208 Commented Jul 31 at 22:10
• BTW, you can't see when a photon hits Joe. You can only see it when it bounces off him and comes back and enters your eyeball. Of course, you can both have cameras (synced to your clocks) that record when you get hit by a photon from the other guy. Commented Jul 31 at 22:17

Here is a great reference that walks you through spacetime diagrams. The majority of questions about SR can be cleared up by seeing the diagram rather than reading words.

I suggest printing out the image right under the heading "Simultaneity," and drawing out your events on the diagram. Stationary observer is the black coordinates and the moving one is the red coordinates. Light rays always travel at 45° on the black coordinates.

After drawing it using the rules described in the article, the answer should become clear.

https://tikz.net/relativity_minkowski_diagram/

You have asked a question that is very similar to one I asked a few weeks ago, In SR, why do we claim length contraction rather than faster than c travel I strongly suggest you look at the question and the answer given by John Rennie.

What it comes down to is this. You might think that you have travelled faster than c by your clock, but the photon will still arrive before you do, so you can then understand that you must NOT have travelled faster than c. The only answer you can come up with is length contraction. It takes a real twist of your thinking to get a grasp of kinetic time dilation from the point of view of the traveler.

• The reason why relativity is so "hard" to understand is merely educational. One can derive the formalism with a few pages of high school algebra from a few assumptions. That is, unfortunately, not what we do when we teach relativity. We are still trying to talk about trains and clocks and smoke and mirrors, which are not exactly clarifying the physics of it. IMHO "the twist" is simply one of a failed education strategy that still prevails for historical reasons. Commented Aug 1 at 20:02
• @foolishmuse You are exactly correct old friend - nothing can outrun photons ever anywhere, in Special or General Relativity. Depending on the coordinate system you can measure your speed anywhere from zero to infinity, but in every single case a photon will beat you to the finish +1. Commented Aug 3 at 4:33
• @FlatterMann as I say, if you understand pythagoras, you can draw spacetime diagrams and understand SR. I think there are people (including here) who thrive on this confusion for their own egos. Clue: endlessly waffling on about relativity of simultaneity is not user-friendly! I think we need to shed that baggage like the ether. Commented Aug 6 at 10:08
• @m4r35n357 The human mind seems to prefer a good mystery over a simple explanation, I guess. In hindsight I would have preferred a much more algebraic introduction to relativity in my own studies. I, too, was brought up with trains and stuff and I couldn't make any sense of it. It's not even what we are observing in reality because we are usually not dealing with synchronized clocks and rulers. We are observing mostly Doppler shifts. Some things like the twin paradox are intuitively far easier to understand in that framework than in Lorentz transformations. Commented Aug 6 at 10:41
• @safesphere unless it is accelerating forever ;) Commented Aug 6 at 10:50

Here's a spacetime diagram drawn on "rotated graph paper" to support the calculations done in the updated-answer by @JEB.

While the choice of relative velocity is good (since its doppler factor is rational and so the light-clock diamonds lie nicely on the rotated grid), the choice of events leads to some fractional results. So, segments don't line up as nicely. (The $$6\times 6$$ subdivision is useful for velocities like (3/5)c and (4/5)c and help visualize the fractional parts.)

(Reminder: the diagonals of a light-clock diamond are Minkowski-orthogonal, and represent the time and space axes of that light-clock.)

For details on this approach, the links provided by @JEB in the comments to the OP are good:
How can time dilation be symmetric? and How to reconcile time dilation equation with concept .
Further details are in my article "Relativity on Rotated Graph Paper" (AJP, 84, 344 (2016)) https://doi.org/10.1119/1.4943251 and an early draft is at https://arxiv.org/abs/1111.7254

So to make this work, you need to follow a standard form for SR problems.

1. Name everyone in alphabetical order: Alice, Bob, Charlie..., in frames $$S$$, $$S'$$, $$S''$$, ...

2. Introduce the "at rest" frame 1st, and so on.

3. Talk about events $$E = [t=0, x=0] = [t'=0, x'=0]'$$ I've decided to use square brackets for events, so they 'pop' visually.

4. Start at the origin: $$[0, 0]$$.

Really, there is no reason to start at $$t=5$$, it's one more detail to cause confusion.

So: Alice fires a photon at:

$$E_0 = [0, 0]$$

Bob (formerly Joe) at this time, in $$S$$ is at:

$$E_1 = [0, 4]$$

an closing at $$\gamma=2$$ (or $$\beta=\frac 1 2 \sqrt 3$$, or $$\omega={\rm arctanh}(\beta)$$...it doesn't matter how you specify it).

....OK, now we see the problem. When doing relativity of dealing with Minkowski diagrams, and Zo is correct: draw a Minkowski diagram, the origins of Alice (you) and Bob (Joe) need to overlap.

So now I need to work backwards, in my head. Thanks.

We need to move the origin up by $$4/({\sqrt 3}/2)= 8/\sqrt{3}$$, which is the event when Alice and Bob are co-located in spacetime.

I quit. Rephrase the question in the frames I have described and the answer will be clear. (Note, if you can't put the problem in this form, it's ill-posed, or it's too complicated to be pedagogical).

Edit: OK, I'm trying again. The definitive answer to:

Is the problem just that we couldn’t start our clocks at the same time?

is "YES". The way the problem is set up, the absolute clock bias is meaningless. Joe and you, as the problem is posed, do not share a common origin event $$O$$:

$$O \equiv [t=0, x=0] \equiv [t'=0, x'=0]'$$

which precludes the use Lorentz transformations.

Since Joe is approaching you, there is an event that is the crossing of your world lines--this should be the origin of $$S$$ and $$S'$$.

The nuisance that led to putting it on hiatus was your choice of $$\beta=\sqrt(3)/2\approx 0.866....$$ it's (mathematically) irrational, and is not nice for a good Minkowski diagram.

If you pump the brakes a bit, and set $$\beta=\frac 4 5=0.8$$, then Pythagorus kicks in and:

$$\gamma = \frac 1 {\sqrt(1-\frac{16}{25})} = \frac 5 3$$

So I'm just going to go with those.

Now I can define coordinates, which you can draw on a Minkowski diagram (exactly if you have 15 sub-tics per integer tic-mark):

You are in $$S$$, and your world line is:

$$W_A(t) = [t, 0]$$

and Joe (aka Bob) is on:

$$W_B(t) = [t, \beta t]$$

This is nice, we have both your worldlines well defined in your coordinates, as a continuous set of events parameterized by your clock.

The question specifies that you fire a photon when $$x_B=-4$$, which we can now read off Joe's worldline:

$$W_B(t, x)_x = \beta t = -4$$,

so we call that event $$E_1$$ and see that occurs at $$t_1 = x_1/\beta = -4/(4/5) = -5$$, making your coordinates of the event:

$$E_1 = [-5, 0]$$

That is you firing a photon in your frame. We can now find where Joe is when that happens (using your definition of simultaneity, which is equal time-coordinates $$\forall x$$): At this time, Joe's world line puts him at:

$$E_2 = W_B(t_1, \beta t_1) = [-5, -4] \equiv [t_2, x_2]$$

(note: integers, nice).

So what does Joe see? Since we have events in co-origin'd frames, we can use the Lorentz Transform:

$$t'_2 = \gamma(t_2-\beta x_2) = \frac 5 3(-5 - \frac 4 5 (-4)) = -3$$

$$x'_2 = \gamma(x_2 - \beta t_2) = \frac 5 3(-4 - \frac 4 5 (-5)) = 0$$

(where the zero is required, since Joe is always at $$x'=0$$).

Thus:

$$E_2 \equiv [t'_2, x'_2]' = [-3, 0]'$$

where the prime, $$)'$$, means the coordinates are in $$S'$$ (I prefer this over saying $$E'_2$$ because events are points in Minkowski space, and they are the same thing in all frames, they just have different coordinates in different frames. Maybe $$\{t, x\}_S$$ and $$\{t', x'\}_{S'}$$ are better--we're trying to maximum clarity--but there is only so much MathJax that I can do).

Meanwhile, we can LT your photon emission into Joe's coordinates:

$$t'_1 = \gamma(t_1-\beta x_1) = \frac 5 3(-5 - \frac 4 5 (-0)) = -\frac{25} 3$$

$$x'_1 = \gamma(x_1 - \beta t_1) = \frac 5 3(0 - \frac 4 5 (-4)) = +\frac{16}3$$

and he puts your transmission at

$$E_1 = [-8 \frac 1 3, +6 \frac 2 3]'$$

Now I hope you see the problem. You simply cannot agree on a synchronous time to emit your photons.

We can define a third event: Joe's position when he sees you transmitting, which is defined by his definition of simultaneity:

$$t'_3 = t'_2 = -\frac{25} 3$$

and of course (which is why we colocated the origins of $$S$$ and $$S'$$):

$$x'_3 = 0$$

so that:

$$E_3 = [-\frac{25} 3, 0]'$$

which we can inverse LT back to you:

$$E_3 = [-\frac{125} 9, \frac{100} 9]$$

which defines a 4th event:

$$E_4 = [-\frac{125} 9, 0]$$

which is YOU when you see that JOE sees YOUR transmission. That is, $$80/9$$ seconds before you transmit, Joe is seeing you transmit, and when you actually transmit, Joe says you transmitted:

$$t'_2-t'_3 = 2\,{\rm seconds}$$

ago. (Note: this is why using events in well defined frames is critical to understanding relativity. Relying on a sentence like "YOU when you see that JOE sees YOUR transmission" stretches the ability of English verb conjugation...it did not evolve with Special Relativity in mind).

There is still a fifth event: when Joe sees your photon, which occurs at

$$t_5 = t_1 + \frac{|x_2|}{\beta+c} = -5 + \frac{4}{9/5} = -\frac{25} 9\,{\rm seconds}$$

so that Joe's position in $$S$$ and $$S'$$ are, respectively:

$$E_5 = [-\frac{25} 9, -\frac{20} 9] = [\frac 5 3, 0]'$$

Now at this point, I'll let you pick a time for Joe to transmit, and see what everyone sees.

The take aways for the OP on physics are:

• Use colocated inertial frames (otherwise, you can't LT)

• Talk about Events (otherwise, it is far to confusing)

• Pick nice $$\beta$$, $$\gamma$$, if possible (seriously, you can see why I bailed at $$\sqrt(3)$$. Pythagorean triples are a good start).

Take Away for OP on PSE:

Take aways for downvoters:

• Don't down vote the only attempt to actually answer the question. New serious students will ask mixed up questions, especially in SR and quantum mechanics (source: https://physics.stackexchange.com), and such questions may not admit a proper answer, so any answer is by definition -1. You can downvote the question, but that is not very inviting to noobs, we don't want this site to go the stack overflow route of gate-keeping code nerds. We're here to help noobs. j/s.
• @m4r35n357 I should have said "put on hiatus". I've answered a lot of SR questions here, and a general theme has arisen: absolutely confusing and scrambled set-ups, and since I want ppl to learn SR, I'm trying to get some ordered approach to solving such problems. But this one is a doozy. Another problem is sub-optimal parameters (even in professional education material). If you use pythagorean triples, you can get $\beta$ and $\gamma$ to be rational, which I think helps--though I am not making progress on this problem.
– JEB
Commented Aug 1 at 16:05
• @FlatterMann I'm showing the thought process. Warts and all. Hopefully it shows how an ill-posed question goes off the rails, and reasoned and deliberate consideration gets it back on track. (and that poor choice of values is problematic...and I am not blaming the OP...$\gamma=2$ seems reasonable. The complaint is more directed at professional works, such as TEDx's youtube.com/watch?v=h8GqaAp3cGs . I mean 23 years? The whole video is sus just based on that choice. It's a prime number, so no $\gamma$ can save the day. You want a base time $T$ such that $T$, $T/\gamma$, $T/\gamma^2$
– JEB
Commented Aug 3 at 5:35
• ....$2T(1-1/\gamma^2)=2T\beta^2$ are rational, if not integer. Why make the problem messier?)
– JEB
Commented Aug 3 at 5:36
• Be careful. In "when you see that JOE sees YOUR transmission", "see" means "the time coordinate of a distant event assigned by an observer [via a spacelike vector (using a line of simultaneity)]". However, in "when Joe sees your photon", "see" refers to receiving a lightlike signal at a local event, on the observer worldline. Commented Aug 5 at 6:10
• @JEB I think it's best to reserve "see" for optical measurements and use instead "measure" or "assign coordinate" for ruler measurements or time-measurements of a distant-event. (For clarity, I prefer thinking in terms of operational radar measurements [Synge,Bondi,Geroch,Ellis-Williams] rather than a lattice of clocks and rulers [Taylor-Wheeler].) Commented Aug 5 at 15:28