Are there surfaces in which clocks in two different inertial frames agree?

This is a homework question I've been working on, and it has been bothering me quite a bit since I'm not sure I understand the essence of the problem. Here it is.

Imagine an inertial frame $S$ and a second inertial frame $S'$ moving relative to $S$ with a constant speed $v$ along only the $x$-axis. Is there a surface in which the clocks in $S$ and $S'$ agree? What about a surface in which the $x$-coordinates agree?

I know that in general, we will see a difference between the two clocks due to time dilation, but the problem asks if there is a special surface in which both agree?

My initial reaction was that, no, this isn't possible, since we would require $\gamma=1$ in order for no time dilation to occur. However, this would imply that the frames are essentially the same, so I don't think this counts. I also had the idea of maybe synchronizing the clocks on a sphere, but then this wouldn't be two inertial frames, since the velocity isn't constant. As such, my thought would be that there is no surface in which this could happen, but I don't know if I'm missing something.

For the part of the question that deals with the $x$-coordinate, once again I feel like we're stuck without an answer, since the question requires one to move along the $x$-axis, so it isn't possible for the spatial $x$-coordinate to be the same in both frames.

Any hint towards working to an answer here would be appreciated. I'm simply looking for a nudge in the right direction.

Update:

I wanted to include a bit more that I've done on the problem. I think the question is whether or not we can have $t'=t$ or $x'=x$, and what surfaces do these imply.

For the first case, if $t'=t$, then using the Lorentz transformation equations gives us:

$$x = ct \sqrt{\frac{\gamma - 1}{\gamma + 1}}.$$

Similarly, if $x'=x$, then we get:

$$x = ct \sqrt{\frac{\gamma + 1}{\gamma - 1}}.$$

I think we can only allow the first case, since the second one implies that $x \gt ct$, which is superluminal. I think that this is the correct answer, though I would appreciate if anyone has more thoughts on the matter.

• Planes normal to the velocity $\:\boldsymbol{\upsilon}\:$ are surfaces of common simultaneity : Two events A,B happening simultaneously in $\:\rm S\:$ on a normal plane are simultaneous in $\:\rm S'\:$ and vice versa. – Frobenius Jan 31 '18 at 22:24

Hint: $$\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$ It is an even function so $-v$ gives the same $\gamma$ as $v$.

Edit: the hint didn't clear any confusion so I will give what I think is the answer. The trick is that you have to introduce a third reference frame. Since $\gamma\geq1$ one of the frames will always see the other one time dilated. If you introduce a new reference frame it is possible that both frames are time dilated by the same amount to the third frame. In one dimension the solution is simple: take your third frame, let's call it $S_0$, to be in the middle of the origins of $S$ and $S'$. That way $S$ and $S'$ will be moving at a speed of $\pm\frac{1}{2}v$ relative to $S_0$. That means $\gamma$ is necessarily the same.

I plotted the one dimensional case in a space-time diagram from the $S_0$ perspective. The coordinates are $(x,ct)$. The thick red/blue lines are the observers $S$ and $S'$ and the normal red/blue lines are the lines of constant time/constant x-coordinate. The observers are moving at $0.15\ c$.

You can see from the diagram that the clocks agree in this frame. When two units of time have passed in the $S$ frame two units of time have also passed in the $S'$ frame. In the $S_0$ frame these events are simultaneous because they are on a horiontal line (the purple line). Meanwhile 2.023 units have passed in the $S_0$ frame, because the purple line intersects at 2.023.

If you looked at the three dimensional case, which points would also have this property of symmetric time dilation?

• Yes, definitely. But how does this give us some kind of surface? Also, wouldn't this still give a time dilation effect, since $\gamma \ne 1$? That's really the issue I'm having. I understand that there are ways to leave the Lorentz factor invariant, but I'm not sure how this can help me find a surface. If there were two frames moving with respect to $S$, I would say that perhaps something like a torus would work, with each additional inertial frame moving with velocity $v$ and $-v$, as you suggested, but I don't know if that's within the scope of the question. – Germ Jan 30 '18 at 23:47
• But the clocks don't agree with one another in the rest frame of either clock. – Chris Jan 30 '18 at 23:54
• Thanks for the edit and the clarification. I definitely see what you're saying. However, I think the question is asking more about if we can construct a surface in which $t'=t$ or $x'=x$, not if we can find a third frame with these properties. If that was the case, you are correct. Also, as @Chris said, in the frame of either clock, I don't think the time would be the same for both of them. It's only when we move out to a suitable frame (as you constructed), that the equivalence occurs. As to your question about the three-dimensional case, I'm going to have to give this some thought. – Germ Jan 31 '18 at 23:44