When I was a little kid, I was fascinated by the fact that we are not able to surpass the speed of light. I imagined a giant spaceship trying to catch a light beam like superman tries to catch flash.

The problem with this thought is that it still remained in my head when I grew up and it really gave me a hard time understanding relativity. And this is due to the fact that the community suggests that the speed of light is the speed limit.

But, is it? It does not make sense to suggest that there is an upper bound to the speed an object can have in the universe. It just doesn't make sense, even if this speed is 1m/s or 100m/s or any arbitrary number. So why the community doesn't suggest that since space and time is the same thing (always in the context of general relativity), there is only one speed in spacetime and this is just an arbitrary speed where every entity in the universe traverses spacetime with this speed? Also, the speed of light maybe should be renamed to the speed between cause and effect.

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    $\begingroup$ Related: physics.stackexchange.com/a/192904/307237 and links therein. "It just doesn't make sense" is another way to reveal that intuition doesn't extend to extreme situations, since our brains were evolved to make sense of the immediate world around us. I would also note that your confusion is immediate to special relativity, see Michelson-Morley. $\endgroup$
    – gmz
    Oct 27, 2021 at 9:42
  • $\begingroup$ @gmz Intuition is what makes us smart actually. If we humans, experience cause and effect with a time delay, this means that things cannot happen immediately other wise the universe would terminate. So what I am trying to suggest is that it would be better if we say from a mathematical perspective that there exists a speed between cause and effect in contrast with this speed is infinite $\endgroup$ Oct 27, 2021 at 9:47
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    $\begingroup$ No, there is an upper bound to how fast objects travel in the universe, with respect to any reference frame. This extends to information, which is why cause-and-effect is finite. Regardless, let us continue this discussion in chat. $\endgroup$
    – gmz
    Oct 27, 2021 at 9:52
  • $\begingroup$ @gmz I don't think that this is a bound. Everyone moves with the speed of cause and effect (or the light speed) but some move in space others in time. But it is not a bound, is the only speed someone can have $\endgroup$ Oct 27, 2021 at 9:54
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    $\begingroup$ Related: Does the speed of light in vacuum define the universal speed limit? and links therein. $\endgroup$
    – Qmechanic
    Oct 27, 2021 at 11:18

3 Answers 3


It does not make sense to suggest that there is an upper bound to the speed an object can have in the universe. It just doesn't make sense, even if this speed is 1m/s or 100m/s or any arbitrary number.

Why doesn't it make sense? The universe is weird, and if we can show experimentally that it behaves in a certain counterintuitive way, we should try to bend our intuition, not remain fixed on it.

After all, what we call intuition developed in a certain "special" environment where certain conditions were always met: for example, things almost never travelled at more than, say, a few tens of meters per second in our ancestral environment.

since space and time is the same thing

This is a bit of an over-simplification: space and time are not the same (neither in special nor in general relativity), although there is a very close connection between them. Heuristically speaking, they can be "transformed into one another" with Lorentz transformations, but we are always able to distinguish them. More technically, the signs of the space and time components in a diagonalized metric are always opposite.

there is only one speed in the spacetime and this is just an arbitrary speed where every entity in the universe traverses spacetime with this speed

This theory is quite easily falsified: we observe things moving at different speeds from one another all the time. Perhaps you mean to use a different definition for the word speed than the usual one (the derivative of position with respect to time), but if so you should clarify what you mean.

Edit: in a comment, OP clarified that they are defining speed between two spacetime events $(x_1, y_1, z_1, t_1)$ and $(x_2, y_2, z_2, t_2)$ as the ratio $$ \frac{(x_1, y_1, z_1, t_1) - (x_2, y_2, z_2, t_2)}{(x_1, y_1, z_1, t_1) - (x_2, y_2, z_2, t_2)} = 1\,. $$

Even without addressing the issue of defining the ratio between two vectors, it seems quite clear that if there is a way to define it, the ratio of something to itself should always be 1.

So, this alternative definition of speed will always yield 1. In a way, this already indicates why it is not typically used as a definition: it assigns 1 to any path, so computing its value does not really serve any purpose.

I am going to speculate regarding the reason why OP thinks this should be the definition of speed, as opposed to something like $$ \frac{(x_1, y_1, z_1) - (x_2, y_2, z_3)}{t_1 - t_2}\,: $$ this seems not to jive with the idea that we are working in spacetime: we should be treating space and time on equal footing, while this clearly separates them out!

Indeed, in relativity we work with four-vectors which include the time component, and the last equation I wrote is not the definition of speed used in that context, but neither is the one proposed by OP.

Instead, what is used is the four-velocity. For a rather complete discussion do refer to its wikipedia article, but let me try to give an expression for the way this definition looks in the same notation as the other two (in the context of special relativity, flat spacetime --- GR complicates things, and is not really illuminating in this context): $$ \frac{(x_1, y_1, z_1, t_1) - (x_2, y_2, z_2, t_2)}{\sqrt{(t_1 - t_2)^2 - (x_1 -x_2)^2- (y_1 -y_2)^2- (z_1 -z_2)^2}}\,. $$

This still does not treat space and time on equal footing, which is about what I was mentioning earlier; the signs in the square root on the bottom are opposite, which allows us to compute what is called the proper time.

A special case of this definition is what happens when we are looking at a stationary object, such that $(x_1, y_1, z_1) =(x_2, y_2, z_2) $. Then, the root on the denominator simply becomes $\left|t_1 - t_2\right|$, and the result we get for the velocity is $(0, 0, 0, 1)$. (I should have mentioned: I am using units where $c=1$).

This definition seems to assign a stationary object "the speed of light in the time direction".

Let us consider a different case: an object moving one $x$ unit every two $t$ units, so that $t_1 - t_2 = 2 ( x_1 - x_2)$. Then, compacting the notation a bit, we get: $$ \frac{(\Delta x, \Delta t)}{\sqrt{\Delta t^2 - \Delta x^2}} = \frac{(\Delta x, 2\Delta x)}{\sqrt{4\Delta x^2 - \Delta x^2}} = \left(\frac{1}{\sqrt{3}}, \frac{2}{\sqrt{3}}\right) $$ so this means moving at "$2 / \sqrt{3} \approx 1.15$ times the speed of light in the time direction", and "$1 / \sqrt{3} \approx 0.58$ times the speed of light in the $x$ direction".

This all should not be taken too literally. It is a convenient different definition for velocity, but it does not neatly map to the intuitive meaning of the word "velocity".

There is also still a concept to discuss: "speed" usually means the magnitude of the velocity vector, but what does "magnitude" mean for four-dimensional vectors? It turns out that computing it like $|v| = \sqrt{v_x^2 + v_y^2 + v_z^2 + v_t^2}$ does not give a meaningful result; instead, the interesting quantity to compute is $|v| = \sqrt{-v_x^2 - v_y^2 - v_z^2 + v_t^2}$.

You can check that this yields 1 (so, the speed of light) for both of our aforementioned examples. Now, this looks quite similar to the initial claim by OP: "everything travels at the speed of light", but now we have the context to interpret it. It does not mean that everything travels at the same speed, but instead, it tells us that the "relativistic velocity vector" we have defined is always normalized.

It can be useful to think of this in terms of degrees of freedom: the regular, non-relativistic velocity has three; one would think that maybe relativistic velocity has four but this is not the case, since the extra degree of freedom is always fixed by the normalization of the four-velocity.

the speed of light maybe should be renamed to the speed between cause and effect

Well, is it the same thing? If I clap my hands and then you hear it, cause has been followed by effect at the speed of sound, which is different from the speed of light. In fact, effect can follow cause at basically any speed slower than the speed of light. Perhaps what you are getting at is the fact that the speed of light is this upper bound. If so, this is a rather uncontroversial point: the name "speed of light" is conventional, but in studying relativity it becomes quite clear that light is not what intrinsically defines it; instead, it is the speed at which all massless particles propagate, and light is just a very well-known example.

Also, all massive particles propagate slower than this, so it is indeed a bound for the speed at which anything can propagate, and therefore also cause anything else to happen.

  • $\begingroup$ Yes, probably speed was not a right choice. But, I think it makes more sense to suggest that everyone travels at the speed of light either in space or time and no one can escape from this intrinsic property, rather than saying you cannot surpass light in space. The latter one is half the truth. $\endgroup$ Oct 27, 2021 at 10:01
  • $\begingroup$ Regardless of its name, how are you defining the property you refer to as "speed"? Since it is non-standard, we should try to agree on the meaning of the term at least, before arguing about its properties. $\endgroup$ Oct 27, 2021 at 10:04
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    $\begingroup$ speed = ((x1, y1, z1, t1) - (x2, y2, z2, t2)) / ((x1, y1, z1, t1) - (x2, y2, z2, t2)) = 1 $\endgroup$ Oct 27, 2021 at 10:06
  • $\begingroup$ I see! now this is more interesting. I'll edit my answer to address this. $\endgroup$ Oct 27, 2021 at 10:09

In relativity, it’s sometimes useful to talk about the four-velocity

$$ \mathbf u = (\gamma c, \gamma \vec u) $$

where $\vec u$ is the more familiar three-dimensional velocity, and $\gamma = 1/{\sqrt{1-u^2/c^2}}$ is the usual Lorentz factor.

Like all four-vectors, the magnitude of this four-velocity is preserved under boosts. An object at rest has four-velocity $\mathbf u = (c, \vec 0)$. Look at that object from a moving frame and its four-velocity has magnitude $\mathbf u^2 = (\gamma c)^2 - (\gamma \vec u)^2 = c^2$. So it’s common to find, in introductory texts on relativity, an invitation to think of the four-velocity as the “speed” of an object’s “motion through spacetime.” An object at rest is “moving through time” with speed $c$, and an object in motion is “moving through spacetime” along a four-dimensional path, at the same rate.

However, this argument gets a little squashy if you try to extend it to light. The four-velocity is a timelike vector whose magnitude is adjusted by $\gamma$ to account for time dilation of moving objects. But the interval between events along the path of a light ray is “light-like,” with relativistic interval zero. There’s no Lorentz transformation which turns a timelike vector into a light-like one. One approach is to imagine “speeding up” an object until its velocity approaches $c$, which gives you

$$ \lim_{|u|\to c} \mathbf u^2 = \lim_{|u|\to c} \frac{c^2-u^2}{1-u^2/c^2} = c^2 \frac 00 = \text{tread carefully} $$

For an object whose mass is identically zero, this limit doesn’t apply, $\gamma$ is undefined in all reference frames, and the “four-velocity” should be null instead of timelike. It would be consistent with experimental evidence if the photon has a very small but nonzero mass. (The current upper limit is $10^{-18}\rm\,eV$.) However we have lots of reasons to believe that photons are fundamentally different from massive particles, with masses that are identically zero.

  • $\begingroup$ Really nice answer. Can you please elaborate on this "An object at rest is “moving through time” with speed c"? $\endgroup$ Oct 28, 2021 at 16:42
  • $\begingroup$ I just meant that "moving through time with speed $c$" is a not-unreasonable description of the timelike four-velocity $\mathbf u = (c,\vec 0)$. I am reluctant to pluck that clause from the caveats and qualifiers that surround it. $\endgroup$
    – rob
    Oct 28, 2021 at 17:32

[...] Also, the speed of light maybe should be renamed to the speed between cause and effect.

In the given context (namely the foundations of geometry and kinematics within the theory of relativity) it is certainly correct, and also didactically advantageous, to think of the relevant speed as signal front speed; cmp. https://en.wikipedia.org/wiki/Front_velocity .

(Speaking in this context instead of "the speed of electro-magnetic waves, in vacuum", for instance, has tradition, but would obiously presume suitable notions and models of "vacuum" and of "electro-magnetism".)

And importantly, of course: Considering signal fronts, i.e. always the very first notion that participants may have of signal indications stated by others (at least in though-experimental principle; disregarding "practical issues" such as finite thresholds or finite resolution) is as far as possible unambiguous and therefore not at all arbitrary but instead highly significant already in defining what we mean by (i.e. how we'd measure) "speed";
and even before that: "distance" (namely according to its so-called chronometric definition, as J. L. Synge noted);
and even before that: whether some signal source and some signal receiver were and remained "at rest wrt. each other" (a.k.a. both members of the same "inertal system"; a.k.a. both "sitting still in space relative to each other.", as W. Rindler has put it.

As a consequence, the "speed" value which would accordingly be determined of the front of any signal being exchanged (by a source and a receiver "sitting still wrt. each other") is necessarily always determined as the signal front speed; and any other measured speed values, e.g. of some thing that's being propagated from source to receiver, are necessarily as multiple or (rather) fraction of the signal front speed (between this source and receiver). It is therefore convenient to abbreviate "the signal front speed" symbolically; as $c_0$, or for short as $c$.


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