I'm aware that distance and displacement, both are independent of reference frames, when the two frames are stationary wrt each other. Because the actual distance (or the shortest distance) between two points remain the same, as long as the two frames are stationary wrt each other, i.e there is no relative velocity between the two frames (Am I correct?)

But what if the other frame starts moving wrt the other? From what I understand, in this case, displacement and distance are no longer frame independent. Are they? Please do let me know.

My question is, are distance and displacement frame independent, always? Even when there's a relative motion between the two frames?

  • $\begingroup$ Are you asking this taking into account special relativity or ignoring special relativity? $\endgroup$ Commented Oct 1, 2018 at 15:01
  • 1
    $\begingroup$ @TrevorKafka: I don't think it matters whether the OP is talking about Galilean relativity or special relativity. I don't think they're asking about special-relativistic length contraction, just about the fact that the distance between events is frame-dependent, which is true in Galilean relativity as well as SR. $\endgroup$
    – user4552
    Commented Oct 1, 2018 at 15:15
  • $\begingroup$ @TrevorKafka I'm still at school, haven't studied special relativity yet. Though I know what things like length contraction and time dilation are, have done a bit of self study out of curiosity. But I guess that's not enough. So I'm just asking in general, are distance and displacement still frame independent if the two frames in question have a relative velocity wrt each other? $\endgroup$
    – 4d_
    Commented Oct 1, 2018 at 15:36

4 Answers 4


The following discussion disregards effects of special relativity and discusses purely Galilean transformations.

If two frames are at rest with respect to each other and oriented parallel to one another, then the transformation between one frame $S$ with coordinates $(x,y,z)$ to the other $S'$ with coordinates $(x',y',z')$ will take the following form, where $x_0, y_0, z_0$ are constants.

$$x' = x + x_0$$

$$y' = y + y_0$$

$$z' = z + z_0$$

Thus, the displacements between two events in $S'$, event 1 $(x_i', y_i', z_i', t_i)$ and event 2 $(x_f', y_f', z_f', t_f)$, in the new frame are given as follows.

$$\Delta x' = x'_f - x_i' = x_f + x_0 - (x_i + x_0) = x_f - x_i = \Delta x$$

$$\Delta y' = y'_f - y_i' = y_f + y_0 - (y_i + y_0) = y_f - y_i = \Delta y$$

$$\Delta z' = z'_f - z_i' = z_f + z_0 - (z_i + z_0) = z_f - z_i = \Delta z$$

Thus, when frames are at rest with respect to each other, distances are preserved. This is true regardless of whether the initial and final positions are measured simultaneously ($t_i = t_f$) or not ($t_i \ne t_f$).

If the frames are moving with respect to each other with a velocity $\mathbf v = (v_x, v_y, v_z)$ and the frames are coincident at $t = 0$, then the transformation between $S$ and $S'$ is as follows.

$$x' = x + v_x t$$

$$y' = y + v_y t$$

$$z' = z + v_z t$$

If you're looking for the distance between the same two events described above in $S'$, we would have the following.

$$\Delta x' = x'_f - x'_i = x_f + v_x t_f - (x_i + v_x t_i) = x_f - x_i + v_x (t_f - t_i) = \Delta x + v_x (t_f - t_i)$$

$$\Delta y' = y'_f - y'_i = y_f + v_y t_f - (y_i + v_y t_i) = y_f - y_i + v_y (t_f - t_i) = \Delta y + v_y (t_f - t_i)$$

$$\Delta z' = z'_f - z'_i = z_f + v_z t_f - (z_i + v_z t_i) = z_f - z_i + v_z (t_f - t_i) = \Delta z + v_z (t_f - t_i)$$

Thus, when frames are moving with respect to each other, distances are preserved only when the events are simultaneous ($t_i = t_f$).


Displacement of a body is dependent on the frame

Suppose a ball is moving on the ground. In the frame of ball the displacement of ball is zero and in the frame of ground there is a some finite displacement

Distance between two points or the shortest distance between two points is not dependent on frame even if those points are moving. Distance between two points will always be same as seen by any observer


Yes distance and displacement depends on frame of referance but distance between two objects is independent of frame of referance .


Distance and displacement are both frame-dependent. For example, if I go to Gettysburg, Pennsylvania, I can say that I'm at the same place where the Civil War battle occurred, but in a different frame of reference I can say that the earth has moved since 1863.

  • $\begingroup$ Are you sure displacement and distance depend on reference frames if the two frames in question are stationary? What you just said, I think your position vector is different when observed from different frames, but change in position vector, i.e your displacement vector is the same, right? $\endgroup$
    – 4d_
    Commented Oct 1, 2018 at 13:25
  • $\begingroup$ What sense does this make? The Earth hasn't moved in its rest frame. $\endgroup$
    – safesphere
    Commented Oct 1, 2018 at 15:02
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    $\begingroup$ Are you sure displacement and distance depend on reference frames if the two frames in question are stationary? Do you mean stationary relative to each other? Frames can't be classified as stationary or not stationary without reference to some other frame. When we say "frame-dependent," we mean that it is possible for the quantity in question to be different in different frames, not that it is always different. Of course there will be special cases where the quantity is the same. Here, that applies when they're at rest relative to one another. $\endgroup$
    – user4552
    Commented Oct 1, 2018 at 15:12
  • $\begingroup$ @safesphere: I don't understand your comment. Could you explain more what you have in mind? I agree that the earth hasn't moved in its own rest frame. $\endgroup$
    – user4552
    Commented Oct 1, 2018 at 15:13
  • $\begingroup$ @BenCrowell I mean, both frames are stationary wrt Earth's reference frame. In this case, it's obvious that distance and displacement are independent, isn't it? And in the second case, both frames have a relative velocity wrt earth other. What I understand is, in this case, distance and displacement are no longer independent of frames. Can you please tell me if I'm correct? $\endgroup$
    – 4d_
    Commented Oct 1, 2018 at 15:32

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