# Why are objects at rest in motion through spacetime at the speed of light?

I read that an object at rest has such a stupendous amount of energy, $E=mc^2$ because it's effectively in motion through space-time at the speed of light and it's traveling through the time dimension of space-time at 1 second per second as time goes forward.

What troubles me here, is the fact that it is traveling through space-time at the speed of light. Why is it at the speed of light?

First, the fact that an object at rest has energy $mc^2$ is a simple matter of dimensional analysis. If you accept that energy and mass are related, and you know that nature has a natural velocity $c$, then $E=mc^2$ is the simplest thing you can write that describes this. The only complication could have been some numerical factor in front of $m$.

Now, the statement about traveling through time 'at the speed of light' needs to be qualified. You can easily see that it does not make sense if you use ordinary definitions: the speed of light is measured in 'length per time', while a 'speed through time' would be measured by 'time per time', which is just a number.

However, we can make sense of this statement. We think of an observer as tracing a path through spacetime. To denote a point on this path we use a single coordinate that we call $\tau$. The path is defined by the functions $t(\tau)$ and $x(\tau)$: for each value of $\tau$ the observer is at a specific place $x$ at a specific time $t$.

Note that so far $\tau$ is not time: it just just a fictitious coordinate that we use to denote points on the path. I did not even have to specify what units we measure it in.

Let's say we measure $\tau$ in seconds. We can now define a 'velocity vector' $u$ through spacetime, which is the rate at which $t$ and $x$ change when we change $\tau$: $$u=\left(c \frac{dt}{d\tau},\frac{dx}{d\tau}\right).$$ Notice that I snuck a $c$ in there, to make sure $u$ has units of velocity.

The first component of $u$ is a nice definition of our velocity through time. The second component is some way of measuring velocity through space, but it is not the same as the velocity we usually think about, which is $\frac{dx}{dt}$.

Now comes a very nice mathematical theorem, which says that we can always assign values of $\tau$ to points on the path such that $|u(\tau)|=c$ at each point. With this choice of $\tau$, our 'velocity' $u$ is a constant, equal to the speed of light: no matter if we still or move very fast, our velocity through spacetime is the same. If we are sitting still, it just means that we are 'moving faster' in the time direction. If we are moving very fast (by the usual definition of moving...), then our velocity in the time direction will be smaller to compensate. Our only choice is where to point our velocity: a bit more in the time direction, or a bit more in the space direction.

I think this is a very beautiful picture, but it is also a bit misleading. Because $\tau$ is a fictitious coordinate, there are many choices for it and they are all equally good. We could just as well have chosen $|u(\tau)|=2c$, or $|u(\tau)|$ not even constant. So bear in mind that this is all a matter of convention.

• Though "arc length traced out along the curve" depends only on the curve and the underlying geometry, and has the nice physical interpretation of "time, as measured by someone moving along the curve, if that curve is timelike" (or length along the curve, if the curve is spacelike). – Jerry Schirmer Oct 2 '12 at 2:09
• Now comes a very nice mathematical theorem, which says that we can always assign values of τ to points on the path such that |u(τ)|=c at each point. Not true. This fails for a massless particle. – user4552 Sep 3 '14 at 23:39
• "If we are moving very fast, then our velocity in the time direction will be smaller to compensate" What this statement says about light which already moves at the highest speed possible ? Does light moves through space only or time only ? or is my assumption wrong – Hammar Feb 12 '17 at 13:59
• The first paragraph is just flat-out wrong. "Nature has a natural velocity" is unsatisfying and not a logical argument -- if you didn't know better, energy could very well be infinity times $m$, like in Newtonian mechanics, or mass may not even have energy. $E=m$ (without any scale factors) is REQUIRED as a result of energy conservation, and the argument isn't so trivial. – Abhimanyu Pallavi Sudhir Jun 7 '18 at 12:15
• The paragraph says "If ... you know that nature has a natural velocity c". This is clearly presented as an assumption, not an argument. We are allowed to make this assumption because it is backed by experiment. – Guy Gur-Ari Jun 7 '18 at 15:29

I read that an object at rest has such a stupendous amount of energy, $E=mc^2$ because it's effectively in motion through space-time at the speed of light and it's traveling through the time dimension of space-time at 1 second per second as time goes forward.

This is wrong.

What troubles me here, is the fact that it is traveling through space-time at the speed of light. Why is it at the speed of light?

It isn't. This idea seems to be something that the popularizer Brian Greene has perpetrated on the world. Objects don't move through spacetime. Objects move through space. If you depict an object in spacetime, you have a world-line. The world-line doesn't move through spacetime, it simply extends across spacetime.

Greene's portrayal of this seems to come from his feeling that because the magnitude of a massive particle's velocity four-vector is traditionally normalized to have magnitude $c$, it makes sense to describe the particle, to a nonmathematical audience, as "moving through spacetime" at $c$. This is simply inaccurate. A good way to see that it's inaccurate is to note that a ray of light doesn't even have a four-vector that can be normalized in this way. Any tangent vector to the world-line of a ray of light has a magnitude of zero, so you can't scale it up or down to make it have a magnitude of $c$. For consistency, Greene would presumably have to say that a ray of light "moves through spacetime" at a speed of zero, which is obviously pretty silly.

The reason we normalize velocity four-vectors for massive particles is that the length of a tangent vector has no compelling physical interpretation. Any two tangent vectors that are parallel represent a particle moving through space with the same velocity. Since the length doesn't matter, we might as well arbitrarily set it to some value. We might was well set it to 1, which is of course the value of $c$ in relativistic units. But this normalization is optional in all cases, and impossible for massless particles.

• Of course the timeline doesn't move. It's stationary. But the object corresponding to this worldline does move through spacetime (along the timeline, which in my view doesn't pre-exist). Do you believe, like Einstein and others did, that the distinction between the past, present and future is a persistent illusion? – descheleschilder Jun 10 '18 at 12:51
• Photons move through space only, non-moving particles only through time. – descheleschilder Feb 18 at 18:48
• Objects move through time too. Why not? ** Greene would presumably have to say that a ray of light "moves through spacetime" at a speed of zero, which is obviously pretty silly.** Sounds pretty silly to me... – descheleschilder Feb 18 at 19:52
• Isn’t normalizing to c special? Isn’t that the length you get using arc-length (proper time) parameterization? – DPatt Feb 19 at 16:52

An object's total velocity through spacetime, its four-velocity, has a component in each dimension of spacetime (4). So, the four coordinates an object has in spacetime are $$x= \left( \begin{array}{ccc} ct \\ x^{1}(t) \\ x^{2}(t) \\ x^{3}(t)\end{array} \right)$$ For the three dimensions of space, we see that these are just the positions as functions of the time as measured by whatever observer whose reference frame we are operating in. At the top, we see ct rather than t. Why? Well, t is measured in units of time, but we need it in units of length to make any sense out of the concept of spacetime. Alternatively, we can convert our positions into units of time by dividing by c. This is done often in astronomy, because the distances are so large (e.g. light-years, light-seconds, etc.).

Now, in order to find the four-velocity, we differentiate each term with respect to proper time. This is similar to the way you find velocity is classical physics - if the position of an object is given by 2t, then it has a velocity of 2. However, note that we are using proper time, $\tau$. The normal t that has been apearing is the time measured by which ever observer who is recording these coordinates, whereas $\tau$ is the time measured on a clock held by our moving observer. So, to find the four velocity, we use this $$\mathbf U = \frac {dx} {d\tau}$$ Where x is the above vector that contains all of the coordinates. Using the fact that time as measured by our observer who is recording the coordinates is related to the proper time of the moving observer by the Lorentz factor, we can write $ct = \gamma c\tau$ Differentiating this with respect to $\tau$, we find that the velocity in the time direction is $c\gamma$. If our 'moving' observer is at rest with respect to the one taking the measurements, then $\gamma = 1$. So, the velocity in the time direction equals c, the speed of light.

We see that for a moving observer, $\gamma$ is greater than one, resulting in a larger velocity through time. This may seem counter-intuitive, since moving observers elapse less time. However, an intuitive way to think about it is this - velocity through space is how muc distance you can cover in some amount of time. So, velocity through time is the amount of time you can elapse in someone's reference frame in some amount of proper time. An observer moving at an enormous speed will record very little proper time, but an observer who he is moving with respect to will observe that it takes an enormous time for him to finish his journey. So, in a small amount of proper time our observer 'travelled' a very large distance through the coordinate time of the observer taking the measurements, hence a higher velocity through time.

Yes, the time-like component of the four-velocity of a stationary object is $c$. It's also correct that "everything moves at the speed of light through spacetime" -- this just means the magnitude of the four-velocity is $c$ (or rather, 1), and its direction keeps changing (note that the transformation here isn't really a rotation, it's a skew/boost, because of how the Minkowski dot product is calculated) as the vector slides on an invariant hyperbola (much like rotations slide on an invariant circle). This is not a matter of convention -- yes, you can choose other parameterisations for the worldline, but they don't satisfy the nice property of becoming equal to co-ordinate time when $v=0$.

Note that the other claim -- that the rest energy is a kinetic energy through time -- is wrong. This is trivially obvious, considering that everything moves at $c$ through spacetime, and the kinetic energy associated with a spatial speed of $c$ is infinity.

Kinetic energy is not best interpreted as "the energy of motion" in special relativity, but instead to be part of the momentum in the time direction. This helps you realise kinetic energy doesn't break the symmetry between space and time at all, it's just that the time-momentum gets increased by a quantity called "kinetic energy" upon a Lorentz boost.

• Yes, the time-like component of the four-velocity of a stationary object is c. This is a statement of normalization, which only applies to massive particles and is in any case an arbitrary assumption. Even assuming this convention about normalization, this doesn't mean that "objects at rest are in motion through spacetime at the speed of light." – user4552 Nov 11 '19 at 19:52
• I did say that it's a convention about normalisation ("yes, you can choose other parameterisations"). I'm saying it's not an arbitrary convention, in the sense that it's perfectly sensible to ask that $\tau$ becomes $t$ for a stationary object. – Abhimanyu Pallavi Sudhir Nov 12 '19 at 11:06
• (note that the transformation here isn't really a rotation, it's a skew/boost, because of how the Minkowski dot product is calculated) as the vector slides on an invariant hyperbola (much like rotations slide on an invariant circle). Nicely put! – descheleschilder Feb 18 at 20:04

The velocity of a mass through spacetime is

$${\eta}^{\mu}=\frac{dx^{\mu}}{d \tau}$$

where $$d\tau=\frac{dt}{\gamma}$$, and, of course, $$\gamma=\frac {1}{{\sqrt{1-{\frac{v^2}{c^2}}}}}.$$ We use $$d\tau$$ instead of $$dt$$, because this is an invariant w.r.t. Lorenz transformations.

Now,

$$\eta^0=\frac{dx^0}{d\tau}=\gamma c,$$

so

$$\eta^{\mu}=\gamma(c,v_x,v_y,v_z).$$

It follows that

$$\eta_{\mu}\eta^{\mu}={\gamma}^2(c^2-{v_x}^2-{v_y}^2-{v_z}^2)=c^2,$$

an invariant. So the absolute value of $$\eta^{\mu}$$ is always equal to c. If the mass has zero (spatial) velocity (the mass is standing still) the speed through time is $$c$$, and photons traveling through space only with speed $$c$$. In between these two extremes, the four-velocity consists of a part moving through time and a part moving through space.

As explained in the former answer is seems that the speed of time is greater when the mass travels through space, but this is due to using $$d\tau$$.

• This is circular reasoning and doesn't answer the question. You state that $\eta^0=\gamma c$, but this assumes that the velocity vector is normalized to $c$, which is what is presumably meant by the claim that "objects are in motion through spacetime at the speed of light." – user4552 Nov 11 '19 at 19:50
• Why doesn't ${\eta}^{0}$ follows from $\eta^0=\frac{dx^0}{d\tau}=\gamma c,$? You can also use $c=1$ (in which case you the percentages of $c$ for going through space and time), or any value of c, but then you have to work this back to c=c as it were to express $c$ in m/sec. I can't see the circularity. – descheleschilder Nov 14 '19 at 5:21