I have came across a few works in gravitational physics using the term "DC" without further explanation of its meaning. For example, consider Strominger's 1703.05448, which states in p. 2 about the memory effect that

This is a subtle DC effect, in which the passage of gravitational waves produces a permanent shift in the relative positions of a pair of inertial detectors.

And a bit later, in the same page,

The former is a statement about momentum space poles in scattering amplitudes, while the latter concerns a DC shift in asymptotic data between late and early times.

In p. 78, we once again have

Integrating this equation reveals a DC effect.

On a completely different topic, consider the reference gr-qc/9301015 by Penrose, Sorkin, and Woolgar (PSW hereafter). In p. 12 it is said that

The function $m^{\mu\nu}$ describes what will be called the DC part of the asymptotic metric

and on the following page we get

They are weak enough to encompass, for example, the DC fields and radiation emitted by any astrophysically plausible source we know of.

A Living Review by Luc Blanchet says, in p. 21, that

It is therefore essentially a zero-frequency effect (or DC effect), which has rather poor observational consequences in the case of the LIGO-VIRGO detectors [...]

So what precisely is the meaning of "DC" in gravitational physics? From the above excerpts, especially the last one, I am guessing "DC" is jargon for "zero frequency", in reference to the idea of "direct current" having no oscillations in electromagnetism. So maybe the use by PSW means that the fields and terms they are characterizing as "DC" are the non-oscillatory contributions, such as the Coulomb-like term of the gravitational field.

Is this understanding precise? In more generality, what is the meaning of "DC" in gravitational physics?

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    $\begingroup$ This usage of "DC" to mean "zero frequency" is not specific to gravitational physics. It is standard jargon throughout physics. $\endgroup$
    – d_b
    Commented Jun 18 at 20:44
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    $\begingroup$ @d_b Good to know. I had never seen it before outside circuits in elementary E&M or in these gravitational contexts $\endgroup$ Commented Jun 18 at 21:36

2 Answers 2


What does "DC" mean...

"DC" means "Direct Current." It is often used in contrast to "AC," which means "Alternating Current."

Alternating current changes with time and is associated with a frequncy (e.g., 60 Hz). Direct current is (in principle) not changing with time.

Therefore, DC can be taken to mean zero frequency, or perhaps very low frequency, very long time period, or very long wavelength.


In this specific context, DC means as other people pointed out "Direct Current." This jargon refers to the $f = 0$ component of a Fourier transform.

If you look at the definition of FT $$\tilde{x}(f) = \int_{-\infty}^{\infty} \mathrm{d} t \, x(t) e^{- i \, 2 \pi f t}$$ where all the frequency dependency is at the exponential, the "DC component" is simply $$ \tilde{x}(f=0) = \int_{-\infty}^{\infty} \mathrm{d} t \, x(t) $$ which is essentially the area under your signal. For finite signals with oscillations around 0, this term is usually pretty small.

The case you are interested in, gravitational memory, looks like an off-set that grows asymptotically towards a non-zero value (see e.g. Fig. 2 of https://arxiv.org/abs/2405.08868. The wiggles on the top left figure die out at zero; the wiggles on the top right column converge into a positive value). Hence, if one were to take their FT, the memory contribution would pile up at low frequencies, especially at $f=0$ for a long-enough signal.

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    $\begingroup$ The existing answer proves that, to explain that DC means the zero frequency component, there is absolutely no reason for overcomplicating. Neither integration, infinity, derivatives, complex number exponents, nor Fourier transform are needed to explain the concept of DC. It looks like this post includes a salad of complicated symbols just for the sake of being different enough from existing answer. $\endgroup$
    – user402514
    Commented Jun 20 at 0:09
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    $\begingroup$ @user402514 The 'salad' you refer to is standard undergraduate mathematics. If you don't know what a Fourier transform is, you're not going to understand how the analogue applies to phenomena not related to electric current. $\endgroup$
    – messenger
    Commented Jun 20 at 4:17
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    $\begingroup$ @messenger If you know what a Fourier transform is, you do not have to see its definition again. If you do not knot what it is, the definition on its own will not help you. $\endgroup$ Commented Jun 20 at 13:43

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