# Why is $dt/d\tau=\gamma$? What is $dt/d\tau$ supposed to mean exactly?

I'm a math student trying to learn some physics by reading Susskind's The Theoretical Minimum. In the volume on special relativity he derives that $$\frac{dt}{d\tau}=\gamma=1/\sqrt{1-v^2}$$ and uses it in the definition of 4 velocity and the action on a particle.

The problem is I don't understand what $$\frac{dt}{d\tau}$$, or rather $$\frac{d\tau}{dt}$$ is supposed to mean mathematically. My first impression was that we have a time-like curve $$C$$ in Minkowski space parametrized by $$t$$ $$C(t)=(t,x(t),y(t),z(t))$$ and that $$\tau$$ is the na function of $$t$$ defined by $$\tau(t)=\sqrt{t^2-x^2-y^2-z^2}$$ and because the curve is time like this would be invertible, and so $$\frac{dt}{d\tau}$$ would just be the derivative of that.

So I'm OK until the step $$\frac{dt}{d\tau}= \frac{1}{\frac{d\tau}{dt}},$$ but he derives $$\frac{d\tau}{dt}=\sqrt{1-v^2}$$ (where I assume $$v$$ is just the time derivative of $$(x(t),y(t),z(t))$$) using some shaky argument: $$d\tau=\sqrt{dt^2-dx^2-dy^2-dz^2}$$ and the divides by $$dt$$.

What is this supposed to mean mathematically? My interpretation must be different from what he had in mind because if we simply differentiate $$\tau(t)=\sqrt{t^2-x^2-y^2-z^2}$$ by $$t$$, using the chain rule we get $$\frac{d\tau}{dt}=\frac{t-xv_1-yv_2-zv_3}{\sqrt{t^2-x^2-y^2-z^2}}$$ Which is different from $$\sqrt{1-v^2}$$ as can be seen by the case when $$v=0$$ and $$x$$ is not.

If we interpret the $$dt$$-s as differential forms with $$d\tau$$ the derivative of $$\tau$$ defined on an open subset on the Minkowski space we get the same result and $$d\tau=\sqrt{dt^2-dx^2-dy^2-dz^2}$$ simply does not hold.

If we just define $$d\tau$$ as function by the above equation the argument would go through, but then what would $$d\tau$$ have to do with the original definition of $$\tau$$? $$d\tau$$ wouldn't even be a 1-form, just some function on the tangent bundle.

I don't understand.

• I don't understand your question. $d\tau^2 = \left(1- \left|\frac{d\vec{x}}{dt}\right|^2\right)dt^2$, that's all. In particular, I don't understand what your function $\tau(t)$ is supposed to mean... Commented May 7, 2023 at 16:01
• Related : Transformation of 4− velocity. Commented May 7, 2023 at 17:05

$$d\tau= \sqrt{dt^2-dx^2-dy^2-dz^2}$$ is the infinitesimal increment of proper time $$\tau$$ along a timelike trajectory $$(x(t),y(t),z(t))$$ parametrized by the coordinate time $$t$$. This is standard calculus 101, so I do not understand you claim that this "simply does not hold". Now we have $${d\tau}= \sqrt{1- \left(\frac{dx}{dt}\right)^2-\left(\frac{dy}{dt}\right)^2-\left(\frac{dz}{dt}\right)^2}dt \\ =\sqrt{1- |v|^2}dt.$$ Thus the elapsed proper time (the Minkowski analogue of arc-length) along the curve is $$\tau(t) = \int^t_0 \sqrt{1- |v(t')|^2}dt'$$

This relates the $$t$$ parameter to the $$\tau$$ parameter. As the curve is everywhere timelike, the map is invertible and $$t$$ can be recovered from $$\tau$$.

Any stuff on differential forms and open sets is just overthinking.

Would you have the same problem with the formula $$s = \int \sqrt{1+ \left(\frac {dx}{dz}\right)^2 + \left(\frac {dy}{dz}\right)^2 } dz$$ computing the arc-length of a nearly vertically-directed Euclidean curve $$(x(z),y(z),z)$$?

• Ok thanks, I understand now. So firstly that last formula is not a theorem, it's the definition of arc-length. (Or rather easily derived from a more general definition for rectifiable curves, but they definitely don't teach that in calc 101) So if we similarly take the first integral to be the definition of elapsed proper time and assume our curve is parametrized by that, then yes everything works out fine. Commented May 7, 2023 at 17:19
• We learned the arc-length formula in high school when we did intrinsic coordinates. We derived it from Pythagoras theorem, so I don't think it can be a definition--- but, of course, back in school we were doing calculs rather than analysis, so all curves were rectifiable. Commented May 7, 2023 at 18:15

It might be helpful to think in terms of geometry and trigonometry using "rapidity" (the Minkowski analogue of the angle between vectors), which is defined between two future-timelike vectors.

With $$v=\tanh\theta$$ (velocity is slope on the position-vs-time graph), we have
$$\frac{dt}{d\tau}=\gamma=1/\sqrt{1-v^2}=\cosh\theta$$.

The time-dilation factor is the [hyperbolic-]cosine of the [Minkowski-]angle between two [future unit-timelike] vectors.

For two 4-velocities $$\hat U$$ and $$\hat W$$, the Minkowski-dot-product is $$\hat U\cdot \hat W=\gamma =\cosh\theta$$.

Then, you can proceed the generalize the worldline of one object to be a more general curve, rather than a line.

I am going to somewhat repeat mike stone's answer with a tiny twist. But in general, if you are even interested in rigour, mathematical or not, avoid Susskind's stuff as much as you can. His theoretical minimum is sloppy, even to physicists!

When you see, in physics, random floating $$\mathrm dx$$ that is neither integrated over nor is a real one-form, often the obvious way to salvage them is to insert an arbitrary division on your own. For example, the obvious way to see that the given relation is correct, is to see that it can be taken as shorthand for $$\frac{\mathrm d\tau}{\mathrm d\lambda}=\sqrt{ \left(\frac{\mathrm dt}{\mathrm d\lambda}\right)^2 -\left(\frac{\mathrm dx}{\mathrm d\lambda}\right)^2 -\left(\frac{\mathrm dy}{\mathrm d\lambda}\right)^2 -\left(\frac{\mathrm dz}{\mathrm d\lambda}\right)^2}$$ for any convenient parameter $$\lambda$$, and if you take $$\lambda=t+c$$ you can see that $$\frac{\mathrm d\tau}{\mathrm dt}=\sqrt{1-\vec v\,^2}$$ as is so important.

Of course, due to the appearance of the square root and squaring, the relationship is not linear, and thus they are all not really one-forms. Arc length type integrands are not one-forms, but after some manipulations, you can find some one-forms (and one-densities) that agree everywhere along some curve, and then you can use the mathematical apparatus of differential forms.

I hope you have the difference between coördinate time $$t$$ and proper time $$\tau$$ conceptually sorted clearly.

There exists a clear contradiction between the relation for Lorentz factor of special relativity and in Lorentz transformation matrix. In special relativity, Lorentz factor is the ratio between coordinate time and proper time whose result is inverse of speed of light and has the dimensions sm^(-1) or second per meter γ = dt/dτ = 1/c (1) while the relation for Lorentz factor in Lorentz matrix is dimensionless γ = Coshθ = 1/(√(1 - v^2/c^2 )) (2) comparing (1) and (2), implies 1/c = 1/(√(1 - v^2/c^2 )) (3) The left-hand side has dimensions while right hand side is dimensionless. This inconsistency has led special relativity to many confusing results in the definition of inner product of 4-vectors where temporal part of inner product retains and spatial part of inner product is omitted.

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• The ratio of two times (presumably measured by clocks) is dimensionless. Commented yesterday

4-vectors and tensors do not remain same in their original form after transformation under Lorentz transformation. All the 4-vectors and tensors based on single transformation law for 4-vectors and tensors (including inner product of 4-vectors and tensors) remain same same for all observers under universal Lorentz transformation matrix. The link of research paper is given here http://pubs.sciepub.com/ijp/11/1/2/index.html

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• Dear Naveed Hussain. Welcome to Phys.SE. For your information, Physics.SE has a policy that it is OK to cite oneself, but it should be stated clearly and explicitly in the answer itself, not in attached links. Commented 23 hours ago